Question

1. solve and graph the inequality 6x² - 7x - 20 < 0

Explanation:

Step1: Change inequality to equation and factor

Set $6x^{2}-7x - 20=0$. Using the AC - method, for $ax^{2}+bx + c$ where $a = 6$, $b=-7$, $c = - 20$, $ac=6\times(-20)=-120$. We need two numbers that multiply to $-120$ and add up to $-7$. The numbers are $-15$ and $8$. Rewrite the middle term: $6x^{2}-15x+8x - 20 = 0$. Group the terms: $(6x^{2}-15x)+(8x - 20)=0$. Factor out the greatest - common factor from each group: $3x(2x - 5)+4(2x - 5)=0$. Then $(3x + 4)(2x - 5)=0$.

Step2: Solve for x

Set each factor equal to zero.
For $3x+4 = 0$, we get $3x=-4$, so $x=-\frac{4}{3}$.
For $2x - 5 = 0$, we get $2x=5$, so $x=\frac{5}{2}$.

Step3: Test intervals

The critical points $x =-\frac{4}{3}$ and $x=\frac{5}{2}$ divide the number line into three intervals: $(-\infty,-\frac{4}{3})$, $(-\frac{4}{3},\frac{5}{2})$, and $(\frac{5}{2},\infty)$.
Test a value from each interval in the original inequality $6x^{2}-7x - 20\lt0$.
For the interval $(-\infty,-\frac{4}{3})$, let $x=-2$. Then $6(-2)^{2}-7(-2)-20=6\times4 + 14-20=24 + 14-20=18\gt0$.
For the interval $(-\frac{4}{3},\frac{5}{2})$, let $x = 0$. Then $6(0)^{2}-7(0)-20=-20\lt0$.
For the interval $(\frac{5}{2},\infty)$, let $x = 3$. Then $6(3)^{2}-7(3)-20=6\times9-21 - 20=54-21 - 20 = 13\gt0$.

Answer:

The solution of the inequality $6x^{2}-7x - 20\lt0$ is $-\frac{4}{3}\lt x\lt\frac{5}{2}$, or in interval notation, $(-\frac{4}{3},\frac{5}{2})$. To graph, draw an open circle at $x =-\frac{4}{3}$ and $x=\frac{5}{2}$ on the number - line and shade the region between them.